What is the tangent of f(x)=5-2x^2,(-2,-3) ?

The tangent of f(x)=5-2x^2,(-2,-3) is y=8x+13

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tangent of f(x)=5-2x^2,(-2,-3)

tangent of

f (x) = 5 - 2 x^{2}, (- 2, - 3)

y = 8 x + 13

Solution steps

Find the slope of

f (x) = 5 - 2 x^{2} : \frac{df ( x )}{d x} = - 4 x

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 8

Find the line with slope m=

8

and passing through

(- 2, - 3) : y = 8 x + 13

y = 8 x + 13

x \to \infty lim (- 2 x + \frac{1}{x})

- 3 cos (5 t)

\frac{x - 5}{x - 2}

\int_{1}^{\infty} x^{- \frac{5}{2}} d x

\frac{\partial}{\partial x} ((2 x - 2)^{4} (x^{2} + x + 1)^{5})

What is the tangent of f(x)=5-2x^2,(-2,-3) ?
The tangent of f(x)=5-2x^2,(-2,-3) is y=8x+13